Is my answer about the Laurent's series correct?

Question

Suppose I have this question:

Find the Laurent's series of $\displaystyle -\frac{4}{3(z+3)^2}$ for $|z|<1$.

This is my answer: $$-\frac{4}{27} \cdot [(1 - z + z^2)]^2.$$

Answer 1

A Laurent series of a complex function $f$ with respect to a particular point $z_0$ is a representation of $f$ as a power series which includes terms of negative degree: $$f(z)=\sum_{n=-\infty}^{+\infty}a_n(z-z_0)^n$$ which holds in some annulus $r<|z-z_0|<R$.

In your case $$f(z)=-\frac{4}{3(z+3)^2}.$$ I guess that you are interested in the point $z_0=0$. Then for $|z|<3$ you should expect something like that $$f(z)=-\frac{4}{27}\cdot\frac{1}{(1+\frac{z}{3})^2}=-\frac{4}{27}\left(1-2\left(\frac{z}{3}\right)+3\left(\frac{z}{3}\right)^2-4\left(\frac{z}{3}\right)^3+\dots\right).$$

P.S. Note that for $|w|<1$, $$\frac{1}{1-w}=\sum_{k=0}^{\infty}{w^k}\implies \frac{1}{(1-w)^2}=\left(\frac{1}{1-w}\right)'=\sum_{k=1}^{\infty}k{w^{k-1}}.$$